Simplify and expand the following expression: $ \dfrac{k}{2k - 6}-\dfrac{k + 6}{3k + 2} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2k - 6)(3k + 2)$ Multiply the first term by $\dfrac{3k + 2}{3k + 2}$ $ \begin{align*} \dfrac{k}{2k - 6} \times \dfrac{3k + 2}{3k + 2} & = \dfrac{(k)(3k + 2)}{(2k - 6)(3k + 2)} \\ & = \dfrac{3k^2 + 2k}{(2k - 6)(3k + 2)}\end{align*} $ Multiply the second term by $\dfrac{2k - 6}{2k - 6}$ $ \begin{align*} \dfrac{k + 6}{3k + 2} \times \dfrac{2k - 6}{2k - 6} & = \dfrac{(k + 6)(2k - 6)}{(3k + 2)(2k - 6)} \\ & = \dfrac{2k^2 + 6k - 36}{(3k + 2)(2k - 6)}\end{align*} $ Now we have: $ = \dfrac{3k^2 + 2k}{(2k - 6)(3k + 2)} - \dfrac{2k^2 + 6k - 36}{(3k + 2)(2k - 6)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{3k^2 + 2k - (2k^2 + 6k - 36)}{(2k - 6)(3k + 2)} $ $ = \dfrac{3k^2 + 2k - 2k^2 - 6k + 36}{(2k - 6)(3k + 2)} $ $ = \dfrac{k^2 - 4k + 36}{(2k - 6)(3k + 2)}$ Expand the denominator: $ = \dfrac{k^2 - 4k + 36}{6k^2 - 14k - 12}$